H.T. Banks and Marie Davidiandavidian/stma810c/lectures/c...References [Andrews]J. F. Andrews, A...
Transcript of H.T. Banks and Marie Davidiandavidian/stma810c/lectures/c...References [Andrews]J. F. Andrews, A...
Introduction to the Chemostat
H.T. Banks and Marie Davidian
MA-ST 810
Fall, 2009
North Carolina State University
Raleigh, NC 27695
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The System
Figure 1: Chemostat schematic.
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We consider the problem of growth of
micro-organisms, for example, a population of
bacteria requiring an energy source containing
carbon for growth–(say a simple sugar). Suppose
we have some bacteria in a container, and we
add nutrients continuously in this container (i.e.,
a continuous culture medium). Assume the
bacteria’s growth depends on a limiting nutrient
alone (i.e., all other nutrients are in excess and
other conditions necessary for their growth are
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adequate). The container has an outlet so that
nutrients and bacteria in the container can flow
out. We further assume the container is well
mixed. Question is how do we understand the
dynamics in the container sufficiently well so as
to operate continuously at equilibrium or steady
state and what are the steady states, if any?
From 1950’s, has lead to many investigations
[BC, Cap, CN, HHW, Monod, NS, Rub, SW, TL]
on modeling of chemostat problems.
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Chemostats are also used as microcosms in
ecology [BHMJA, PK] and evolutionary biology
[WMB, DD, WWE, JE] as well as in wastewater
treatment based on chemostat models [BHWW]
and has led to numerous patents!!!! . In the some
cases, mutation/selection is detrimental, in other
cases, it is the desired process under study.
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Chemostat dynamics and their understanding
have led to many mathematical investigations
including those of Wolkowicz and co-authors –
see [BW] and 30+ subsequent references–
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The Mathematical Model
Modeling is based on compartmental analysis,
laws of mass action, and mass balance. To
formalize the problem, we introduce some
notation.
∙ Let V be the volume of the chemostat (the
container, and it is fixed in our example) in
the unit of liters (l).
∙ Let Q be the volumetric flow rate (flow rate
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into and out of the chemostat) in unit of liters
per hour (l/ℎr).
∙ Let q = QV be the dilution rate in the unit of 1
per hour (1/ℎr); (then 1q is the mean residence
time for a particle in the growth chamber).
∙ Let N(t) be the mass of bacteria at time t in
the unit of gram (g), N(t0) = N0.
∙ Let c(t) be the concentration of rate limiting
nutrient in the unit of gram per liters (g/l),
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c(t0) = 0.
∙ Let r be the growth rate of the bacteria in
units of per hour (1/ℎr). Assume growth is
enzyme mediated (e.g., growth of E. coli with
nutrient galactose via enzyme
galactosekinase). Then one expects r is a
function of c ( as we will explain below, we
will use Michaelis-Menten/ Briggs-Haldane
kinetics for saturation limited growth rates).
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We can describe the chemostat with a coupled
set of differential equations derived using mass
balance and laws of mass action: (dmdt ∝ �(m)) :
dN(t)dt = r(c(t))N(t)− qN(t)
dc(t)dt = qc0 − qc(t)− 1
yr(c(t))N(t).
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Here the term qc0 represents the input rate, i.e.,
the rate we add nutrients into the container and
y is the yield parameter where y ∝ Y , where Y is
the yield constant defined by
Y =mass of bacteria change/time
mass of nutrient consumed/time.
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Typically, r(c) is assumed to have the form
r(c) =Rmaxc
Km + c
. So that r(c) can not exceed Rmax and it will
approach Rmax when c→∞, i.e., saturation
limited kinetics. This is based on
Michaelis-Menten/Briggs-Haldane reaction
kinetics [BanksLN, BH, MM, Rub]–more later.
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Steady State of the System
We are interested in operating the chemostat
under steady state or equilibrium conditions.
The chemostat dynamical system for which we
wish to find the steady state is
N(t) = r(c(t))N(t)− qN(t)
c(t) = q(c0 − c(t))−1
yr(c(t))N(t),
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where
q =Q
Vand r(c) =
Rmaxc
Km + c.
For this system, we want to find the steady state,
i.e., we want to find constants (N , c) such that
dN
dt
∣∣∣∣(N ,c)
= 0dc
dt
∣∣∣∣(N ,c)
= 0.
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Set
r(c)N − qN = 0,
and to obtain cases of interest assume N is
non-zero, so dividing, obtain r(c) = q, can
explicitly solve for c thus:
Rmaxc
Km + c= q
or (Rmax − q)c = qKm and hence
c =Kmq
Rmax − q.
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Next, we set
q(c0 − c)−1
yr(c)N = 0,
but since r(c) = q, we see that
(c0 − c)−1
yN = 0,
which means that
N = y(c0 − c).
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Thus, to sum up, we see that a nontrivial steady
state is given by
(N , c) =
(y(c0 − c),
Kmq
Rmax − q
). (1)
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Thus, when simulating for Homework 1, a good
check to verify the coding has been done
correctly is to run the simulation over a long
time period and verify that it tends to the steady
state. In other words, one should verify that as
t→∞, N(t)→ N and c(t)→ c.
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Remark: There also exists a trivial solution to
the equilibrium problem, specifically
(N , c) = (0, c0), which is easily verified from the
original system. However, this solution is not of
interest to us, as it implies that N = 0, which
implies that there is no action taking place (it
also represents an unstable equilibrium).
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We see from (1) that c depends on c0, which
means for any c0, we will arrive at a different
equilibrium, but if we were to start at (N , c), we
would stay there, and this is true if (N , c) is
(0, c0) or that represented by (1). However, if we
start at a point other than an equilibrium, for
the (1) equilibria, it will converge to those
equilibria, but, as we shall later establish, it will
never converge to (0, c0) because by definition
y ∕= 0, and since we started at a point other than
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equilibrium, c0 ∕= c therefore, N ∕= 0. Thus we
say that (0, c0) is an unstable equilibrium while
the nontrivial state (N , c) of (1) is a stable
equilibrium.
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For a discussion of
Michaelis-Menten/Briggs-Haldane kinetics, we
turn the summary Brief Review of Enzyme
Kinetics, Chapter 1 of [BanksLN] – see also
[Rub].
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